Answer
$10$
Work Step by Step
Area $=(1/2) bh=(1/2) \times (4/5) (5-x) \cdot (3/5) (5-x)$
We integrate the integral to calculate the volume as follows:
$V= \int_{0}^{5} (\dfrac{6}{25}) (5-x)^2 dx$
Now, $V= (-\dfrac{2}{25})(5-x)^3]_{0}^{5}$
or, $= (-\dfrac{2}{25}) (0-125)$
or, $=10$
